Question

How do you factor \[y=6{{x}^{2}}+11x+4\] ?

Answer 1

Hint: From the question given polynomial is \[y=6{{x}^{2}}+11x+4\]. For answering this question, we will use factorization. Factorization is the process of deriving factors of a number which divides the given number evenly. Factorization is the process of reducing the bracket of a quadratic equation, instead of expanding the bracket and converting the equation to a product of factors which cannot be reduced further. There are many methods for the factorization process. Now, we will do the given question by the method of splitting the constant and doing the sum product pattern. Here first we will find the two numbers whose product is \[24\] and whose sum is \[11\] . Then we will equate to the zero and we will find the factors of the given \[y=6{{x}^{2}}+11x+4\]

Complete step by step solution:
Now considering from the question we have an expression \[y=6{{x}^{2}}+11x+4\] for which we need to derive the factors.
We can factor the \[{{x}^{2}}-11x+30\]by below method:
Given equation is in the form of \[a{{x}^{2}}+bx+c=0\]
First, we have to divide \[24\]into the product of the two numbers and Sum of those two numbers must be equal to the coefficient of\[x\]i.e., \[11\]
Now, \[24\] can be split into the product of the two numbers.
\[24\]can be split into product of \[8\]and \[3\]
Their sum is also equal to \[11\] which is equal to the coefficient of\[x\].
\[6{{x}^{2}}+11x+4\]
\[\Rightarrow 6{{x}^{2}}+8x+3x+4=0\]
\[\Rightarrow 2x\left( 3x+4 \right)+3x+4=0\]
\[\Rightarrow \left( 3x+4 \right)\left( 2x+1 \right)=0\]
\[\Rightarrow x=-\dfrac{4}{3}\] and \[x=-\dfrac{1}{2}\]
Therefore\[x=-\dfrac{4}{3}\],\[x=-\dfrac{1}{2}\] are the factors of \[y=6{{x}^{2}}+11x+4\]

Note: During answering questions of this type, we should be sure with our calculations. For answering this question, we can also use the formulae for obtaining the roots of the quadratic equation$a{{x}^{2}}+bx+c=0$ given as $\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$ then if the two solutions are $p,q$ then the factors will be $\left( x-p \right)\left( x-q \right)$ . For \[y=6{{x}^{2}}+11x+4\] the roots are \[\dfrac{-11\pm \sqrt{121-4\left( 24 \right)}}{2\times 6}=\dfrac{-11\pm 5}{12}=-\dfrac{4}{3},-\dfrac{1}{2}\] then the factors will be \[\left( 3x+4 \right)\] and\[\left( 2x+1 \right)\].